On the double series $sum_{(m,n)inmathbb{Z}^2setminus{(0,0)}}frac{m^2+4mn+n^2}{(m^2+mn+n^2)^s}$

It seems to me that this is a trick question because the series vanishes for reasons of symmetry.

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Let $vec{u}$ and $vec{v}$ be two unit vectors forming a 60 degree angle. It is easy to see (law of cosines) that $$ ||mvec{u}+nvec{v}||^2=m^2+mn+n^2. $$ Let $R$ be the sixty degree rotation, i.e. the linear transformation defined by $R(vec{u})=vec{v}$ and $R(vec{v})=vec{v}-vec{u}$. So $R$ is an order six transformation, and for all vector $vec{w}$ we have $$||R(vec{w})||^2=||vec{w}||^2.$$ Consider an orbit of $langle Rrangle$, that is, fix a pair $(m,n)$, and apply the transformations $R^i, i=0,1,ldots,5$, to the vector $vec{w}_0=mvec{u}+nvec{v}$. It follows that the vectors $vec{w}_i=R^i(vec{w}_0)$, $i=0,1,ldots,5$, all have the same length. In other words, the pairs $$(m,n), (m+n,-m), (n,-m-n), (-m,-n), (-m-n,m), (-n,m+n)$$ all contribute terms sharing the same denominator.

But the numerators of such a sextet of terms add up to zero.

Write $R(m,n)=m^2+4mn+n^2$. Then the first three vectors contribute $$ begin{aligned} &R(m,n)+R(m+n,-m)+R(n,-m-n)\ =&m^2+4mn+n^2+(m+n)^2-4m(m+n)+m^2+n^2-4n(m+n)+(m+n)^2\ =&(1+1-4+1+1)m^2+(4+2-4-4+2)mn+(1+1+1-4+1)n^2\ =&0. end{aligned} $$ The negatives of those three vectors give the same sum for reasons of parity, and the claim follows.

Whenever the series converges absolutely, the derangement theorem allows us to sum it one sextet at a time, and the series vanishes.

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The vectors $vec{w}$ with roughly equal norm $R^2$ fall on the circle of radius $R$ (give or take $1/2$, rounding to the nearest integer here). Given that the vectors form the hexagonal lattice, there are $mathcal{O}(R)$ such vectors (scales with the length of the perimeter of that circle). The absolute values of those terms contribute thus sum up to roughly $mathcal{O}(R^3/R^{2s})$. It is well known that the sum over $R$ converges absolutely, when $s>2$, so we are done.